5.9 Representations of Functions as a Power Series
5.9
Brian E. Veitch
Representations of Functions as a Power Series
Example 5.58. The following geometric series
X
xn = 1 + x + x2 + x3 + x4 + . . .
will converge when −1 < x < 1. We also know that a geometric series with radius x will
converge to
X
xn =
1
1−x
So what did we just do? We found a power series representation for the function
1
f (x) =
. But, this representation only works for values of x where −1 < x < 1.
1−x
Who cares, right? Wrong! We care. Remember that an infinite series can be approxi1
mated by adding up the first N terms. For us, we can approximate f (x) =
by
1−x
1
≈1+x
1−x
1
≈ 1 + x + x2
1−x
1
≈ 1 + x + x2 + x3
1−x
1
≈ 1 + x + x2 + x3 + x4
1−x
All of these approximate our function. However, the more terms we add the better the
approximation. Let’s take a look at the graph of all these.
285
5.9 Representations of Functions as a Power Series
Brian E. Veitch
See? More terms equals better approximation. If I use
1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14
and zoom out alot, we get this
I want to draw your attention to something. When does the approximation begin to
1
. It appears to deviate when |x| ≥ 1. This is exactly when
really deviate from f (x) =
1−x
X
the series
xn diverges. Interesting...
During the last section, we spent
X a great deal of time determining what values of x will
make the following power series
cn (x − a)n converge.
What we’re going to do in this section, is find ways of rewriting some functions so they
can be written like a power series.
Example 5.59.
1. Express
4
as a power series.
1 + x2
286
5.9 Representations of Functions as a Power Series
The trick is rewrite
Brian E. Veitch
P n
4
a
=
ar .
so
that
it
can
look
something
like
1 + x2
1−r
4
4
=
2
1+x
1 − (−x2 )
We can call a = 4 and r = −x2 . Therefore,
X
X
4
2 n
=
4(−x
)
=
4(−1)n x2n
1 + x2
Next, let’s find the interval ov convergence.
(−1)n+1 x2n+2 4(−1)n+1 x2(n+1) = lim 4x2 =
lim
L = lim n→∞
4(−1)n x2n n→∞ (−1)n x2n which converges when
2
4x < 1
Solving for x, we get
2 1
x <
4
|x| <
1
2
1
1
− <x<
2
2
2. Express
2
as a power series.
3−x
Rewrite
2
2
2/3
as
=
3−x
3(1 − x/3)
1 − (x/3)
Let a =
2
x
and r = , we get
3
3
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5.9 Representations of Functions as a Power Series
Brian E. Veitch
∞
X 2 x n−1
2
=
3−x
3 3
1
x
as a power series.
+1
x
as
Rewrite 2
2x + 1
3. Express
2x2
x·
1
1
=
x
·
1 + 2x2
1 − (−2x2 )
If we let r = −2x2 , we get
x
1
= x·
+1
1 − (−2x2 )
∞
X
(−2x2 )n
= x
2x2
0
∞
X
(−2)n x2n
= x
0
=
∞
X
(−2)n x2n+1
0
If we write out the first few terms, it would look like
x
= x − 2x3 + 4x5 − 8x7 + 16x9 − 32x11 + ...
2x2 + 1
And this series will converge as long as |−2x2 | = 2x2 < 1.
We’ve been lucky in that the functions we’re working with start off looking similiar to
a
. Now let’s see what happens when we differentiate or integrate a series. This could
1−r
open up a variety of new functions.
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5.9 Representations of Functions as a Power Series
5.9.1
Brian E. Veitch
Term by Term Differentiation and Integration
Given the function f (x) =
∞
X
cn (x − a)n , whose domain is the interval of convergence.
0
Theorem 5.7. If a power series
f (x) =
∞
X
P
cn (x − a)n has radius R > 0, then
= c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + . . .
0
is differentiable on (a − R, a + R), and
1. f 0 (x) = c1 + 2c2 (x − a) + 3c3 (x − a)2 + 4c4 (x − a)3 + . . . =
Z
2.
f (x) dx = c + c0 (x − a) + c1
Example 5.60. Express
X
ncn (x − a)n−1
X (x − a)n+1
(x − a)3
(x − a)4
(x − a)2
+ c2
+ c3
+ ... = c +
cn
2
3
4
n
1
as a power series.
(1 − x)2
Is there a function we know of that has a power series representation and can be differ1
.
entiated or integrated to give us
(1 − x)2
The answer is yes. Note that
1
d
1
=
dx 1 − x
(1 − x)2
and we know what the power series representation of
1
. Yay!!!
1−x
1
d
1
=
(1 − x)2
dx 1 − x
"∞
#
d X n
=
x
dx n=0
=
=
∞
X
d n
x
dx
n=0
∞
X
nxn−1
n=1
= 1 + 2x + 3x2 + 4x3 + . . .
289
5.9 Representations of Functions as a Power Series
Brian E. Veitch
You may also approach is like this
1
1
d
=
(1 − x)2
dx 1 − x
d 1 + x + x2 + x3 + x4 + . . .
=
dx
= 0 + 1 + 2x + 3x2 + 4x3 + . . .
∞
X
=
nxn−1
n=0
=
∞
X
nxn−1
n=1
Note that the index changed from n = 0 to n = 1. We can do this because the first term
in that series is 0, and therefore don’t need it.
Example 5.61. Find a power series representation for ln(1 + x) and it’s radius of convergence.
Z
We know
1
dx = ln(1 + x). So we figure out the power series representation for
1+x
1
and integrate it to get ln(1 + x).
1+x
1
1
=
1+x
1 − (−x)
∞
X
=
(−x)n
n=0
=
∞
X
n=0
Now let’s integrate back
290
(−1)n xn
5.9 Representations of Functions as a Power Series
Brian E. Veitch
Z
1
dx
1+x
Z X
∞
(−1)n xn dx
=
ln(1 + x) =
=
n=0
∞
XZ
(−1)n xn dx
n=0
=
∞
X
(−1)n
n=0
xn+1
+C
n+1
Use this to find ln(1/2).
Since we found the power series representation for ln(1 + x), we need to figure out what
x is to get ln(1/2). Notice that x = −1/2.
ln(1/2) = ln(1 + (−1/2))
∞
X
(−1/2)n+1
=
(−1)n
n+1
n=0
=
∞
X
(−1)n
n=0
=
=
(−1)n+1 (1/2)n+1
n+1
∞
X
(−1)(1/2)n+1
n=0
∞
X
n=0
n+1
−1
(n + 1)2n+1
Example 5.62. Find a power series representation for f (x) = ln(1 + x), f (x) = x ln(1 + x),
f (x) = ln(x2 + 1), and ln(1 − x).
1. f (x) = ln(1 + x)
Note that
d
1
[ln(1 + x)] =
and we know
dx
1+x
∞
X
1
1
=
=
(−1)n xn
1+x
1 − (−x)
0
291
5.9 Representations of Functions as a Power Series
Brian E. Veitch
Therefore,
Z
ln(1 + x) =
1
dx =
1+x
Z
Z X
X
X
xn+1
n n
n
(−1) x dx =
(−1)
xn dx =
(−1)n
n+1
0
0
0
2. f (x) = x ln(1 + x).
Here, we just multiply our series representation for ln(1 + x) by x.
x ln(1 + x) = x ·
X
(−1)n
0
XX
xn+1
xn+2
=
(−1)n
n+1
n+1
0
3. f (x) = ln(x2 + 1)
This series is similiar to ln(x + 1), except we replaced x with x2 . So all we need to do
is replace x with x2 in our power series representation for ln(x + 1) from part (1).
f (x) = ln(x2 + 1) =
X
(−1)n
0
(x2 )n+1 X
x2n+2
=
(−1)n
n+1
n+1
0
4. f (x) = ln(1 − x)
xn+1
. We just need to replace x with
n+1
0
−x and we’ll have the power series representation for ln(1 − x).
Again, recall that we know ln(1 + x) =
ln(1 − x) =
X
(−1)n
X
(−x)n+1 X
(−1)n+1 xn+1 X xn+1
(−1)n
=
(−1)n
=
−
n
+
1
n
+
1
n+1
0
0
0
Note that (−1)n · (−1)n+1 = −1
Example 5.63. Find a power series representation for f (x) = tan−1 (x).
292
5.9 Representations of Functions as a Power Series
Note that
Brian E. Veitch
d
1
tan−1 (x) =
. Therefore,
dx
1 + x2
Z
1
dx = tan−1 (x)
2
1+x
Now we just need the power series representation for
1
1 + x2
1
1
=
2
1+x
1 − (−x2 )
∞
X
=
(−x2 )n
n=0
=
∞
X
(−1)n x2n
n=0
Therefore,
−1
Z
1
dx
1 + x2
Z X
=
(−1)n x2n dx
tan (x) =
n=0
=
XZ
(−1)n x2n dx
n=0
=
X
(−1)n
n=0
x2n+1
+C
2n + 1
Example 5.64. Find a power series representation for f (x) = tan−1 (x/3).
Like the previous example, we have a power series representation for f (x) = tan−1 (x).
So we just replace the x in tan−1 (x) with (x/3).
tan−1 (x/3) =
X
(x/3)2n+1 X (−1)n x2n+1
(−1)n
=
2n
+
1
32n+1 (2n + 1)
0
0
x
Note the radius of convergence is when < 1.
3
293
5.9 Representations of Functions as a Power Series
Brian E. Veitch
Example 5.65. Suppose I wanted to evaluate the following integral
Z
ln(1 − t5 )
dt
t
This isn’t a very friendly integral (even for calculus II). Instead, we write our integrand
as a power series and integrate that. A power series representation essentially rewrites your
function into a polynomial, which if you remember from calc I, is extremely easy to integrate
and differentiate.
X
Z
ln(1 − t5 )
dt =
t
Z
−
X t5n+5
(t5 )n+1
Z
Z
−
5n+4
n + 1 dt =
n + 1 dt = − X t
dt
t
t
n+1
X Z t5n+4
X
t5n+5
=−
dt = −
n+1
(5n + 5)(n + 1)
So here are the steps to integrating and differentiating complicated functions using a
power series.
1. Figure out how to rewrite your function into a power series. This may take a while
(like the previous example).
2. Use the term by term intergration or differentiation technique we utilized in the previous example.
3. Remember that integral and derivative notation can be pulled in or out of an summation.
4. Finally, clean it up by simplifying.
5. If for some reason, it’s a definite integral (has bounds), then write out the power series
for the first 4 or 5 terms. This way you have a polynomial. Just evaluate the polynomial like you would in any integration problem. Just keep in mind the final answer is
only an estimate. The estimate gets better when you use more terms of your series.
294
5.9 Representations of Functions as a Power Series
Brian E. Veitch
Example 5.66.
Z
.5
.1
ln(1 − t5 )
d
t
(a) Figure out the power series reprsentation for the integrand (which we did already)
and integrate
(b) −
X
0
.5
t5n+5
(5n + 5)(n + 1) .1
(c) Write out a few terms of the series
.5
t5 t10 t15 t20 −
−
− −
5
20
45
80 .1
(.5)5 (.5)10 (.5)15 (.5)20
(.1)5 (.1)10 (.1)15 (.1)20
−
−
−
−
−
−
−
− −
= −0.0062975
5
20
45
80
5
20
45
80
295